1.电容器动态问题分析
2.电场与电场强度
3.库仑定律
4.电路中的电源
5.分压式和限流式对滑动变阻器的要求
6.探究感应电流的产生条件
7.磁通量
8.洛伦兹力与安培力的关系
9.螺旋测微器的原理与读数
10.游标卡尺的原理与读数
11.光电门的原理
12.追及相遇问题
13.小船过河
14.力的平行四边形定则演示实验
15.力的合成
16.动态平衡问题一
17.动态平衡问题二
18.向心加速度的方向
19.万有引力与重力的关系
20. 传送带问题
21.机械振动
22.简谐运动
23.单摆
24.受迫振动
25.共振
26.机械波
27.波的衍射
28.波的干涉
29.静电场
30.静电屏蔽
31.研究电荷分布
32.库仑扭秤实验
33.模拟电场线
34.测量等势线
35.探究电容大小的影响因素
36.氢原子模型
37.带电粒子在电场中的往复运动(一)
38.带电粒子在电场中的往复运动(二)
恒定电流
39.研究路端电压
40.探究欧姆定律
41.探究电阻的伏安特性曲线
42.探究小灯泡伏安特性曲线
43.用DIS研究小灯泡U-I图像
44.用DIS测电源电动势和内阻
45.灵敏电流计改装成电压表
46.半偏法测量电流表内阻
47.磁场
48.电流的磁效应
49.质谱仪
50.回旋粒子加速器
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燕平娜2467螺旋测微器的读数______毫米;游标卡尺读数______毫米 -
项宽段18686337258 ______[答案] 1、螺旋测微器的固定刻度为0.5mm,可动刻度为40.0*0.01mm=0.400mm,所以最终读数为0.5mm+0.400mm=0.900mm.2、游标卡尺的主尺读数为:3cm=30mm,游标尺上第6个刻度和主尺上某一刻度对齐,所以游标读数为6*0.1mm=0....
燕平娜2467螺旋测微器的读数方法有哪些?
项宽段18686337258 ______ 螺旋测微器的读数方法:首先读出固定套筒上所露出来的刻度,然后读出微分同尺寸看清周围情况,要与固定套筒中线基本对齐,然后再乘以0.01毫米,就可以得出微分筒...
燕平娜2467请分别写出游标卡尺和螺旋测微器对应的读数(1)如图1, - -----mm;(2)如图2,------mm -
项宽段18686337258 ______ 游标卡尺的主尺读数为104mm,游标读数为0.05*1=0.05mm,所以最终读数为:104mm+0.05mm=104.05 mm. 螺旋测微器的固定刻度读数为1.5mm.可动刻度读数为0.01*38.0mm=0.380mm,所以最终读数为:1.5mm+0.380mm=1.880mm;故答案为:104.05; 1.880
燕平娜2467如图甲、乙所示是螺旋测微器和游标卡尺测量工件长度时的情形.螺旋测微器读数为______mm.游标卡尺读数为______mm. -
项宽段18686337258 ______[答案] 由图可知,螺旋测微器的固定部分为6.5mm;转动部分为22.3,故读数为:6.5+22.3*0.01=6.723mm; 游标卡尺游标为20分度,长度为19mm;故游标最小分度为 19 20=0.95mm;游标卡尺对齐的主尺长度为37mm,而游标对齐的刻度为:7*0.95=6....
燕平娜2467螺旋测微器的读数 - -----mm(图1),游标卡尺的读数------cm(图2 -
项宽段18686337258 ______ 1、螺旋测微器的固定刻度为4mm,可动刻度为48.5*0.01mm=0.485mm,所以最终读数为4mm+0.485mm=4.485mm. 2、游标卡尺的主尺读数为:2.6cm=26mm,游标尺上第4个刻度和主尺上某一刻度对齐,所以游标读数为4*0.1mm=0.4mm,所以最终读数为:26mm+0.4mm=26.4mm=2.64cm. 故答案为:4.485,2.64
燕平娜2467高中螺旋测微器、游标卡尺等物理仪器的读数方法及精确度.(附例题)谢谢! -
项宽段18686337258 ______ 用游标卡尺?原理1,卡钳是由少量被放大,然后再读取. 2,常用卡钳精确到0.1mm,有的可精确达到0.05mm和0.02毫米. 3,它可以由厚度或物体的外径,该容器的孔或宽度,内径的深度进行测量. 4,游标卡尺读数方法:光标0蜱取其上读取的整个毫米的L1,然后看哪条主标尺刻度上的刻度与主刻度线对齐,读出由以下的小数的L2光标毫米,总读数为:L = L1 + L2.注1,卡钳方法的右手.右前脚的抓地力,用拇指推光标左手按住事情. 2,测量前检查零点. 3.要了解阅读方法.
燕平娜2467读出如图所示中游标卡尺测量的读数和螺旋测微器测量的读数.(1)______cm,(2)______mm,(3)______mm. -
项宽段18686337258 ______[答案] (1)由图示游标卡尺可知,主尺示数为6.1cm,游标尺示数为12*0.05mm=0.60mm=0.060cm, 游标卡尺示数为6.1cm+0.060cm=6.160cm; (2)由图示螺旋测微器可知,固定刻度示数为0.5mm,可动刻度示数为14.1*0.01mm=0.141mm, 螺旋测微器示数...
燕平娜2467如图1所示的螺旋测微器的读数为: - -----mm.如图2所示的游标卡尺的读数为:------mm -
项宽段18686337258 ______ 螺旋测微器的固定刻度读数6mm,可动刻度读数为0.01*12.3=0.123mm,所以最终读数为:固定刻度读数+可动刻度读数为6+0.123=6.123mm...
燕平娜2467高中物理、化学中各种实验仪器的精确度(读数读到哪一位)游标卡尺、螺旋测微器、滴定管、量筒、…………你再据点例子 -
项宽段18686337258 ______[答案] 游标卡尺(0.1mm 0.02mm 0.05mm三种,不估读) 螺旋测微器小数点后三位(0.001m 最后一位是估读的) 滴定管小数点后两位 量筒根据刻度估读一位
燕平娜2467(1)有一段横截面是圆形的金属丝,螺旋测微器测量它的直径如图所示,螺旋测微器的读数为 - -----mm.(2) -
项宽段18686337258 ______ (1):螺旋测微器的读数为:d=4mm+0*0.01mm=4.000mm;(2):①:由于通过待测电阻的最大电流为:I max = U R x =3 5 A=0.6A,所以电流表应选 A 2 ;②:由于待测电阻满足 R V R x > R x R A ,所以电流表应用外接法,把待测电阻与电流表看做一个整体,根据欧姆定律可知,电阻的测量值应小于真实值;③:根据闭合电路欧姆定律可知,电路中需要的最大电阻为:R max = E 1 3 I A =4.5 1 3 *0.6 Ω=22.5Ω,所以变阻器可以采用限流式接法(分压式接法也可以),又电流表应用外接法,电路图如图所示:故答案为:(1)4.000 (2)① A 2 ;②小于;③如图
